Some troubling assumptions of SR

On 6 Mar, 05:47, " wrote:
On Mar 3, 7:41 am, "George Dishman" wrote:
wrote in message ps.com...
....
SR says there is no aberration
so there is no difference in the pointing angle. The predicted
pointing angle is the same but they disagree on whether there is
aberration due to the different definitions of aberration.

OK, but that's a hypothetical case. The Earth is in orbit
and what I am considering is whether a dragged aether can
the explain the actual observed aberration.

Lets try this. You are standing in the open with no wind. An
airplane passes by from left to right. If the airplane dropped a
cannon ball observers on the plane would see the ball drop straight
down, staying directly under the plane as it fell. The observer on
the ground would see the ball dropping from left to right. Following
its path back up leads to a point behind the current location of the
plane, where the plane was when it dropped the ball. This is
aberration. Just consider the airplane to be a stationary star and we
are on the earth moving in our orbit. Have the airplane fly back in
the opposit direction (we have continued our orbit under the
stationary star) and the aberration angle changes direction.

When the cannon ball was dropped a charge of black powder was lit off
with a bang, leaving a cloud of smoke in the stationary air at the
point where the ball was dropped. That is the point we will hear the
sound come from in our stationary air (dragged aether). The airplane
will have moved on in the time it took for the sound to reach us, so
the sound will come from behind the airplane, just like the cannon
ball. When the airplane flys back the other way the sound trails in
the opposite direction.

So far we agree.

If you want the star to be at rest in the aether of space just change
the airplane to a balloon floating back and forth in the jet streams.
It wont effect the final leg of the sound's passage to us in our
stationary air.

That's where we disagree. Rather than the jet stream, suppose
there is a uniform wind at all heights above 100m but the air
in that last 100m is still. There is a shear at that height.
Suppose someone on the balloon fires a gun to create a sound
wavefront.

The sound propagates vertically down through the air from
the balloon as seen by someone in the balloon and the
wavefront is always horizontal:

b
_|_
_|_
_|_
_|_
_|_
.... _|_ ....

From the point of view of someone on the ground, the ballon
and the sound waves are carried sideways by the wind but
the wavefronts remain horizontal. When the sound reaches
the shear, the balloon has drifte to 'b' from the point
where the gun was fired at 'g'


b g
_/_
_/_
_/_
_/_
_/_
.... _/_ ....

After the shear, the sound continues vertically and the
ballon drifts on to 'x' which it reaches when the gunshot
is heard on the ground by observer 'o':

x b g
_/_
_/_
_/_
_/_
_/_
.... _/_ .... shear
_|_
_|_
______o______ ground

The aberration is quite different from an aircraft
in uniformly moving air.

Yes, we use the down wind origin of the wave to calculate where the
wave front will be, but that is not where the sound came from.

It is where it appears to come from. Try drawing circles
radiating out from the source.

I disagree. This is what the real world experiment with sound
demonstrates. When there is a cross wind between two stationary
observers they still hear the sound come from the direction of the
source, not the down wind center of the wave front.

I have yet to be convinced of that. What was the link
to your experimental evidence again?

I provided no link. I was speaking of first hand observations you can
make yourself. Surely you have been in open areas when the wind was
blowing. Have you ever heard come from down wind of a stationary
object?

I have never tried it with equipment capable of measuring
the angle accurately enough and I doubt you have either. If
you have your eyes open, your brain identifies the source
with a combination of what your hear and what you see. Try
it wearing a blindfold and pointing to the source using
sound alone, get a friend to put two stakes in the ground
marking the direction, then take the blindfold off. You
will need a high wind speed to get a measurable offset.

I never have even when the distance was the better part of a
mile and the wind strong.

The distance doesn't matter, the sin of the angle is v/c
where c is the speed of sound.

I have often heard the sound from an
airplane come from a point behind its current location, so I can
detect a difference in direction if one exists.

An aircraft travels at 600 mph, I doubt you have tried
the experiment blindfold is as much as 60 mph when you
would get 1/10th of the displacement.

I have also
experienced traveling at high rates of speed near others at race
tracks.

I wouldn't recommend trying that blindfold ;-)

To determine the direction the sound is coming from the receiver needs
at least two points.

Right. Try this sketch:

S

--- wind

|
A--+--B

A and B are microphones and S is a source, say a gun which
emits a single spherical wavefront, or you can use a tone
generator and measure phase difference.

We are trying to duplicate a wave traveling in the aether. The
signals in the wires travel faster than the sound waves so you are in
effect using faster than light communication to determine the timing
of the reception at A and B.

But in a dragged aether, both the aether at ground level
and the cables are at rest relative to the ground (which
is how dragged aether explains the MMX) therefore the
speed of the signals in the cables is equal. As long as
the cables are the same length, there is no time
difference introduced, the actual speed is irrelevant.

Use a dish to collect the sound at a focal
point.

A and B would then be the surface near the edges of the
parabolic dish.

If aimed at the down wind center the wave front contacts the
up wind and down wind edges of the dish at the same time.

Or put another way the signal reaches the two microphones
simultaneously hence the short line shown normal to the
line joining them would point to the source, downwind.
Alternatively, instead of microphones, think of turning
your head until the sound reached both ears simultaneously.

Imagine synchronizing clocks with sound while assuming there is no
wind. If there is in fact a wind the clocks will be out of sync. But
you have no way of knowing that based on the sound signals alone.

In a dragged aether, there is no wind at ground level,
the wind only exists beyond the interface between the
Earth-dragged and Sun-dragged regions.

But the
down wind path is reflected up wind while the up wind path is
reflected down wind traveling faster in relation to the dish. The two
paths do not converge at the focal point of the dish.

Ah, that adds a second effect. Now suppose the dragged
aether is dragged by the tube of the telescope so there
is no relative motion of the aether between the dish and
the focal point, what do you get?

The quick answer is that it would equalize the times,

Exactly.

but we may be
overlooking something. Does the dragged aether cause a build up of
pressure as it passes the stationary aether?

That would again occur at the boundary between the Earth- and
Sun-dragged regions and we are looking out through the side
where there is a shear.

Do the waves reflect in
the same direction when they hit the dish? Where exactly does the
transition take place and along what path? Instant transitions can
give incorrect results as do instant accelerations.

Think of the transition being something like the
magnetosphere.

Aim the dish a
bit up wind and the down wind path gets shorter while the up wind path
gets longer, bringing the time of the two paths closer to equal. When
aimed at the source the time for the two paths to the focal point is
the same.

Is it? The light spent a long time travelling from
the distant star to the dish and a very short time
from the dish to the focal point. However the angles
between the paths and the aether flow are also greater
so can you show the effects exactly cancel? How would
that affect a VLBI system?

The paths were very nearly the same for the long trip while they were
in opposite directions from the dish to the focal point.

Sure, but the difference is what we are interested in.

Aim further up wind and the down wind path's time becomes
shorter than the up wind path's.

As mentioned previously, if you want to block the sound you place an
obstruction on the line of sight path between the source and
receiver. Placing it between the receiver and the down wind center of
the wave front will have no effect.

Agreed, if you place a block on the line between you and
the source, the sound which apparently comes from somewhere
downwind will be silenced.

Doesn't that seem strange to you?

Lots of things _seem_ strange but science bears them out.
Measure it is the fundamental rule.

In the case of sound coming from
behind an airplane flying by you block the sound by holding something
up in the direction of where the sound is coming from. Blocking the
line of sight does nothing. Why the difference?

In the first case the source is not moving, the two are different
configurations.

.. How can blocking the
line of sight path effect the path from down wind?

Because the sound is actually coming from the object which
hasn't moved but the wavefronts lie at an angle to the line
of sight like a swimmer crossing a river, he has to face
slightly upstream in order to swim perpendicular to the bank.

George

In article .com,
" wrote:

On Mar 4, 8:31 pm, Mitchell Jones wrote:
In article ,
Mitchell Jones wrote:

In article . com,
" wrote:

[snip]





In my previous post we looked at two observers with a cross wind
blowing perpendicular to their line of sight. In the real world if
one of them makes a noise the other will hear it coming straight from
the other reguardless of the wind.

***{True, as I noted earlier in a response to George. --MJ}***

This is true despite the fact that
the center of the circular wave reaching the receiver has drifted down
wind.

***{Yup. If the sound is a backfire from a parked car at A and an
observer is at B, with both points fixed relative to the ground, the
source of the sound in the air mass will be marked by a puff of smoke
that will drift downwind at the wind speed. The wave segment that is
destined to enter the ear of the observer will travel radially outward
from the puff of smoke. The angle between that radius and line AB will
be Tan^-1 (V1/V2)

***{Note: I have been using Tan^-1 (V1/V2), which works well enough as a
rule of thumb, but that's only because at very small angles tan x and
sin x are virtually identical. To be exact, you should use Sin^-1
(V1/V2), as indicated in my 2nd post to this thread (Feb. 17). In the
present case, assuming a 100 mph crosswind and a speed of sound of 770
mph, Tan^-1 (100/770) = 7.4 degrees, and Sin^-1 (100/770) = 7.46
degrees. It generally doesn't matter which you use, as a practical
matter, but in this case, where you are likely focused on the
mathematics as you try to evaluate the point about P moving along AB,
use of the arctangent may lead to confusion.

I would add that it is a simple matter to prove that the true path of
the sound is radially at the aberration angle rather than along the line
AB. If it were along AB, and AB had a length of one mile, the
propagation delay would be (1/770)(3600) = 4.68 sec. In fact, however,
the sound travels down the radius, the length of which is 5280/cos 7.46
= 5325 feet. At 770 mph, or 1129 ft/sec, it will therefore take
[(5325/5280)/770][3600] = 4.72 sec, rather than 4.68 sec, to reach B.

Bottom line: the stronger the crosswind, the greater the true distance
traveled by the sound and the larger the propagation delay; and if the
crosswind is strong enough--i.e., if it is not less than 770 mph--the
sound will never reach the observer at B.

--Mitchell Jones}***

I understand what you are saying, but you have chosen to view the path
of P from someone floating with the air rather than from the
stationary observers. As you said below, P travels along the line
AB.

***{As I said, that is the *apparent* path. The true path is from the
puff of smoke outward along the expanding radius to B. The proof is that
if you use the length of AB as the distance and divide it by the speed
of sound, you will not get the actual propagation delay. With AB being
1 mile and the speed of sound being 770 mph, the calculated delay would
be 1/770th hour, or 4.68 sec. In fact, however, the delay will be the
length of the radial path outward from the puff of smoke, which is
1.0085 miles, divided by 770 mph. That gives .0013 hours, or a 4.72 sec
propagation delay, not 4.68 sec.

Let's be specific about the nature of the disagreement we are having.
You want to treat paths between different "frames" as if they are all
equal. By such a concept, there is no such thing as a "true path."
Please think about what that would mean. Suppose, for example, that the
segment of the circular wave that eventually reaches B were to leave a
trail of smoke as it moved through the air mass. Do you think that trail
of smoke would lie along AB? The answer, of course, is that it would
not. At the instant when the circular wave passed B, the trail of smoke
would lie along the radius connecting the puff of smoke to B.

On the other hand, if we consider the true path of a vibration through
the ground--i.e., a seismic wave--originating from A at the same time as
the sound in the air was made at A, the true path of the seismic wave
segment that arrived at B, in fact, would lie along AB, because the line
AB is fixed relative to the ground. Consideration of such examples leads
us to understand what the concept of a "true path" really means. In
general, the true path taken by an object or phenomenon is the path of
the disturbance that accompanies its motion--which means: it is the
trail left in the medium, i.e., the local material environment, through
which the object or phenomenon is passing.

The true path of a swimmer, for example, is the path of the disturbance
he makes in the water through which he is passing. It is NOT "just as
good" to refer to the path he follows "in the Sun's frame," where he
would have a velocity component of 18 miles/sec, plus a 1000 mph
component due to the rotation of the Earth, in addition to his velocity
component relative to the lake. It would be utterly absurd to say that
the guy is swimming at speeds in excess of 18 miles/sec, and it is
equally wrong to say that the path of the sound in the example we have
been discussing is along AB. We have been talking about an audible sound
passing through air, and there is no disturbance along AB because the
air mass through which the sound propagates is in motion perpendicular
to AB.

The claim that there is no preferred reference frame is tantamount to a
claim that there is no true path taken by an object in motion, and, as
such, is clearly incorrect.

--Mitchell Jones}***

P is the energy that B will eventually detect as sound. The fact
that the air that P was once in has drifted down wind does not change
where P has been when viewed from B. P came straight across from A.
Yes, P had to travel through more air to get to B than it would have
in still air, but that didn't alter its path. If the sound came from
down wind as you say then you would be able to block it by placing
something in the way of the down wind path. But that wont work.
Reality is that you block it by placing something on the line AB.

***{I guess you didn't read my earlier response to that point. Here it
is again: when you place a barrier across AB, you also place a barrier
across the radial path from the puff of smoke. That's because the wave
segment that eventually reaches B is always positioned on the line AB as
the puff of smoke drifts downwind. Thus when the wave segment reaches
the barrier, the line from the puff of smoke to the segment also reaches
the barrier. You can't put a barrier across AB without putting a barrier
across the radial path. In other words, you CAN block the signal by
placing a barrier across the radial path. Hence your argument fails.
--MJ}***

Bruce Richmond

, and the point P on the expanding circular wave which
is destined to reach B will always remain over the line AB as the circle
expands. Why? Because for a radius expanding at that angle, the downwind
velocity of the puff of smoke is exactly canceled by the upwind
component of motion of P. The true path of P (i.e., radially outward
through the air mass from the puff of smoke) will thus deviate from the
apparent path (AB) by the aberration angle. --MJ}***

************************************************** ***************
If I seem to be ignoring you, consider the possibility
that you are in my killfile. --MJ

On Mar 7, 5:15 am, "George Dishman" wrote:
On 6 Mar, 05:47, " wrote:

On Mar 3, 7:41 am, "George Dishman" wrote:
wrote in oglegroups.com...
...
SR says there is no aberration
so there is no difference in the pointing angle. The predicted
pointing angle is the same but they disagree on whether there is
aberration due to the different definitions of aberration.

OK, but that's a hypothetical case. The Earth is in orbit
and what I am considering is whether a dragged aether can
the explain the actual observed aberration.

Lets try this. You are standing in the open with no wind. An
airplane passes by from left to right. If the airplane dropped a
cannon ball observers on the plane would see the ball drop straight
down, staying directly under the plane as it fell. The observer on
the ground would see the ball dropping from left to right. Following
its path back up leads to a point behind the current location of the
plane, where the plane was when it dropped the ball. This is
aberration. Just consider the airplane to be a stationary star and we
are on the earth moving in our orbit. Have the airplane fly back in
the opposit direction (we have continued our orbit under the
stationary star) and the aberration angle changes direction.

When the cannon ball was dropped a charge of black powder was lit off
with a bang, leaving a cloud of smoke in the stationary air at the
point where the ball was dropped. That is the point we will hear the
sound come from in our stationary air (dragged aether). The airplane
will have moved on in the time it took for the sound to reach us, so
the sound will come from behind the airplane, just like the cannon
ball. When the airplane flys back the other way the sound trails in
the opposite direction.

So far we agree.

If you want the star to be at rest in the aether of space just change
the airplane to a balloon floating back and forth in the jet streams.
It wont effect the final leg of the sound's passage to us in our
stationary air.

That's where we disagree. Rather than the jet stream, suppose
there is a uniform wind at all heights above 100m but the air
in that last 100m is still. There is a shear at that height.
Suppose someone on the balloon fires a gun to create a sound
wavefront.

The sound propagates vertically down through the air from
the balloon as seen by someone in the balloon and the
wavefront is always horizontal:

b
_|_
_|_
_|_
_|_
_|_
.... _|_ ....

From the point of view of someone on the ground, the ballon

and the sound waves are carried sideways by the wind but
the wavefronts remain horizontal. When the sound reaches
the shear, the balloon has drifte to 'b' from the point
where the gun was fired at 'g'

b g
_/_
_/_
_/_
_/_
_/_
.... _/_ ....

If there were a stationary observer at the bottom of the diagonal line
watching b drift by, where would he hear the sound come from, b or g ?

After the shear, the sound continues vertically and the
ballon drifts on to 'x' which it reaches when the gunshot
is heard on the ground by observer 'o':

x b g
_/_
_/_
_/_
_/_
_/_
.... _/_ .... shear
_|_
_|_
______o______ ground

The aberration is quite different from an aircraft
in uniformly moving air.

Yes, we use the down wind origin of the wave to calculate where the
wave front will be, but that is not where the sound came from.

It is where it appears to come from. Try drawing circles
radiating out from the source.

I disagree. This is what the real world experiment with sound
demonstrates. When there is a cross wind between two stationary
observers they still hear the sound come from the direction of the
source, not the down wind center of the wave front.

I have yet to be convinced of that. What was the link
to your experimental evidence again?

I provided no link. I was speaking of first hand observations you can
make yourself. Surely you have been in open areas when the wind was
blowing. Have you ever heard come from down wind of a stationary
object?

I have never tried it with equipment capable of measuring
the angle accurately enough and I doubt you have either. If
you have your eyes open, your brain identifies the source
with a combination of what your hear and what you see. Try
it wearing a blindfold and pointing to the source using
sound alone, get a friend to put two stakes in the ground
marking the direction, then take the blindfold off. You
will need a high wind speed to get a measurable offset.

I never have even when the distance was the better part of a
mile and the wind strong.

The distance doesn't matter, the sin of the angle is v/c
where c is the speed of sound.

When you view a car going by 40 feet away the angle defined by the car
is much larger than it is when the car is 4000 feet away. If the car
blows its horn at 40 feet and the sound is shifted back a foot it
would still seem to come from the front of the car. The same angle at
4000 feet would shift the sound by 100 feet, or about 5 car lengths
behind the car.

I have often heard the sound from an
airplane come from a point behind its current location, so I can
detect a difference in direction if one exists.

An aircraft travels at 600 mph, I doubt you have tried
the experiment blindfold is as much as 60 mph when you
would get 1/10th of the displacement.

I have easily detected the shift with an old prop driven plane flying
at ~100 mph. I can also detect it when cars drive by at ~60 mph if I
am 100 yards or more from the highway. When a jet is traveling at 600
mph the effect is so great that it sometimes takes a few seconds to
find the jet.

I have also
experienced traveling at high rates of speed near others at race
tracks.

I wouldn't recommend trying that blindfold ;-)

To determine the direction the sound is coming from the receiver needs
at least two points.

Right. Try this sketch:

S

--- wind

|
A--+--B

A and B are microphones and S is a source, say a gun which
emits a single spherical wavefront, or you can use a tone
generator and measure phase difference.

We are trying to duplicate a wave traveling in the aether. The
signals in the wires travel faster than the sound waves so you are in
effect using faster than light communication to determine the timing
of the reception at A and B.

But in a dragged aether, both the aether at ground level
and the cables are at rest relative to the ground (which
is how dragged aether explains the MMX) therefore the
speed of the signals in the cables is equal. As long as
the cables are the same length, there is no time
difference introduced, the actual speed is irrelevant.

Use a dish to collect the sound at a focal
point.

A and B would then be the surface near the edges of the
parabolic dish.

If aimed at the down wind center the wave front contacts the
up wind and down wind edges of the dish at the same time.

Or put another way the signal reaches the two microphones
simultaneously hence the short line shown normal to the
line joining them would point to the source, downwind.
Alternatively, instead of microphones, think of turning
your head until the sound reached both ears simultaneously.

Imagine synchronizing clocks with sound while assuming there is no
wind. If there is in fact a wind the clocks will be out of sync. But
you have no way of knowing that based on the sound signals alone.

In a dragged aether, there is no wind at ground level,
the wind only exists beyond the interface between the
Earth-dragged and Sun-dragged regions.

But the
down wind path is reflected up wind while the up wind path is
reflected down wind traveling faster in relation to the dish. The two
paths do not converge at the focal point of the dish.

Ah, that adds a second effect. Now suppose the dragged
aether is dragged by the tube of the telescope so there
is no relative motion of the aether between the dish and
the focal point, what do you get?

The quick answer is that it would equalize the times,

Exactly.

but we may be
overlooking something. Does the dragged aether cause a build up of
pressure as it passes the stationary aether?

That would again occur at the boundary between the Earth- and
Sun-dragged regions and we are looking out through the side
where there is a shear.

Do the waves reflect in
the same direction when they hit the dish? Where exactly does the
transition take place and along what path? Instant transitions can
give incorrect results as do instant accelerations.

Think of the transition being something like the
magnetosphere.

Aim the dish a
bit up wind and the down wind path gets shorter while the up wind path
gets longer, bringing the time of the two paths closer to equal. When
aimed at the source the time for the two paths to the focal point is
the same.

Is it? The light spent a long time travelling from
the distant star to the dish and a very short time
from the dish to the focal point. However the angles
between the paths and the aether flow are also greater
so can you show the effects exactly cancel? How would
that affect a VLBI system?

The paths were very nearly the same for the long trip while they were
in opposite directions from the dish to the focal point.

Sure, but the difference is what we are interested in.

Aim further up wind and the down wind path's time becomes
shorter than the up wind path's.

As mentioned previously, if you want to block the sound you place an
obstruction on the line of sight path between the source and
receiver. Placing it between the receiver and the down wind center of
the wave front will have no effect.

Agreed, if you place a block on the line between you and
the source, the sound which apparently comes from somewhere
downwind will be silenced.

Doesn't that seem strange to you?

Lots of things _seem_ strange but science bears them out.
Measure it is the fundamental rule.

In the case of sound coming from
behind an airplane flying by you block the sound by holding something
up in the direction of where the sound is coming from. Blocking the
line of sight does nothing. Why the difference?

In the first case the source is not moving, the two are different
configurations.

.. How can blocking the
line of sight path effect the path from down wind?

Because the sound is actually coming from the object which
hasn't moved but the wavefronts lie at an angle to the line
of sight like a swimmer crossing a river, he has to face
slightly upstream in order to swim perpendicular to the bank.

George

When that swimmer reaches you, after leaving the bank at a point
straight across, do you claim that he swam up from some point down
stream because that is where the water is that he swam through?

Bruce

On Mar 8, 1:04 pm, Mitchell Jones wrote:
In article .com,





" wrote:
On Mar 4, 8:31 pm, Mitchell Jones wrote:
In article ,
Mitchell Jones wrote:

In article . com,
" wrote:

[snip]

In my previous post we looked at two observers with a cross wind
blowing perpendicular to their line of sight. In the real world if
one of them makes a noise the other will hear it coming straight from
the other reguardless of the wind.

***{True, as I noted earlier in a response to George. --MJ}***

This is true despite the fact that
the center of the circular wave reaching the receiver has drifted down
wind.

***{Yup. If the sound is a backfire from a parked car at A and an
observer is at B, with both points fixed relative to the ground, the
source of the sound in the air mass will be marked by a puff of smoke
that will drift downwind at the wind speed. The wave segment that is
destined to enter the ear of the observer will travel radially outward
from the puff of smoke. The angle between that radius and line AB will
be Tan^-1 (V1/V2)

***{Note: I have been using Tan^-1 (V1/V2), which works well enough as a
rule of thumb, but that's only because at very small angles tan x and
sin x are virtually identical. To be exact, you should use Sin^-1
(V1/V2), as indicated in my 2nd post to this thread (Feb. 17). In the
present case, assuming a 100 mph crosswind and a speed of sound of 770
mph, Tan^-1 (100/770) = 7.4 degrees, and Sin^-1 (100/770) = 7.46
degrees. It generally doesn't matter which you use, as a practical
matter, but in this case, where you are likely focused on the
mathematics as you try to evaluate the point about P moving along AB,
use of the arctangent may lead to confusion.

I would add that it is a simple matter to prove that the true path of
the sound is radially at the aberration angle rather than along the line
AB. If it were along AB, and AB had a length of one mile, the
propagation delay would be (1/770)(3600) = 4.68 sec. In fact, however,
the sound travels down the radius, the length of which is 5280/cos 7.46
= 5325 feet. At 770 mph, or 1129 ft/sec, it will therefore take
[(5325/5280)/770][3600] = 4.72 sec, rather than 4.68 sec, to reach B.

Bottom line: the stronger the crosswind, the greater the true distance
traveled by the sound and the larger the propagation delay; and if the
crosswind is strong enough--i.e., if it is not less than 770 mph--the
sound will never reach the observer at B.

--Mitchell Jones}***

I understand what you are saying, but you have chosen to view the path
of P from someone floating with the air rather than from the
stationary observers. As you said below, P travels along the line
AB.

***{As I said, that is the *apparent* path. The true path is from the
puff of smoke outward along the expanding radius to B. The proof is that
if you use the length of AB as the distance and divide it by the speed
of sound, you will not get the actual propagation delay. With AB being
1 mile and the speed of sound being 770 mph, the calculated delay would
be 1/770th hour, or 4.68 sec. In fact, however, the delay will be the
length of the radial path outward from the puff of smoke, which is
1.0085 miles, divided by 770 mph. That gives .0013 hours, or a 4.72 sec
propagation delay, not 4.68 sec.

If a swimmer swims across a flowing river from A to B by swimming at
an angle so that he stays on the line AB he will have swam further
than if there were no current and it will take him longer to get to
B. But that is not reason enough to claim that he really swam to B
from point C somewhere down stream, where the water is that was at A
when he started his swim.

Let's be specific about the nature of the disagreement we are having.
You want to treat paths between different "frames" as if they are all
equal. By such a concept, there is no such thing as a "true path."
Please think about what that would mean. Suppose, for example, that the
segment of the circular wave that eventually reaches B were to leave a
trail of smoke as it moved through the air mass. Do you think that trail
of smoke would lie along AB? The answer, of course, is that it would
not. At the instant when the circular wave passed B, the trail of smoke
would lie along the radius connecting the puff of smoke to B.

At any given instant the intersection of the wave front with line AB
is a point. Once it has past through a volume of air it doesn't
matter what happens to that air. That packet of energy is no
different than a bullet. Someone floating by in a balloon would see
it pass by on a diagonal, but A and B see its path as the straight
line between them. If we want to know what A and B will see we have
to view things from the perspective of A and B.

On the other hand, if we consider the true path of a vibration through
the ground--i.e., a seismic wave--originating from A at the same time as
the sound in the air was made at A, the true path of the seismic wave
segment that arrived at B, in fact, would lie along AB, because the line
AB is fixed relative to the ground. Consideration of such examples leads
us to understand what the concept of a "true path" really means. In
general, the true path taken by an object or phenomenon is the path of
the disturbance that accompanies its motion--which means: it is the
trail left in the medium, i.e., the local material environment, through
which the object or phenomenon is passing.

So when a bullet is fired from A to B just as a balloon floated by A,
is the true path AB or CB?

The true path of a swimmer, for example, is the path of the disturbance
he makes in the water through which he is passing. It is NOT "just as
good" to refer to the path he follows "in the Sun's frame," where he
would have a velocity component of 18 miles/sec, plus a 1000 mph
component due to the rotation of the Earth, in addition to his velocity
component relative to the lake. It would be utterly absurd to say that
the guy is swimming at speeds in excess of 18 miles/sec, and it is
equally wrong to say that the path of the sound in the example we have
been discussing is along AB. We have been talking about an audible sound
passing through air, and there is no disturbance along AB because the
air mass through which the sound propagates is in motion perpendicular
to AB.

The claim that there is no preferred reference frame is tantamount to a
claim that there is no true path taken by an object in motion, and, as
such, is clearly incorrect.

--Mitchell Jones}***

P is the energy that B will eventually detect as sound. The fact
that the air that P was once in has drifted down wind does not change
where P has been when viewed from B. P came straight across from A.
Yes, P had to travel through more air to get to B than it would have
in still air, but that didn't alter its path. If the sound came from
down wind as you say then you would be able to block it by placing
something in the way of the down wind path. But that wont work.
Reality is that you block it by placing something on the line AB.

***{I guess you didn't read my earlier response to that point. Here it
is again: when you place a barrier across AB, you also place a barrier
across the radial path from the puff of smoke. That's because the wave
segment that eventually reaches B is always positioned on the line AB as
the puff of smoke drifts downwind. Thus when the wave segment reaches
the barrier, the line from the puff of smoke to the segment also reaches
the barrier. You can't put a barrier across AB without putting a barrier
across the radial path. In other words, you CAN block the signal by
placing a barrier across the radial path. Hence your argument fails.
--MJ}***

Bruce Richmond

, and the point P on the expanding circular wave which
is destined to reach B will always remain over the line AB as the circle
expands. Why? Because for a radius expanding at that angle, the downwind
velocity of the puff of smoke is exactly canceled by the upwind
component of motion of P. The true path of P (i.e., radially outward
through the air mass from the puff of smoke) will thus deviate from the
apparent path (AB) by the aberration angle. --MJ}***

************************************************** ***************
If I seem to be ignoring you, consider the possibility
that you are in my killfile. --MJ- Hide quoted text -

- Show quoted text -

On 9 Mar, 02:03, " wrote:
On Mar 7, 5:15 am, "George Dishman" wrote:
On 6 Mar, 05:47, " wrote:
On Mar 3, 7:41 am, "George Dishman" wrote:
wrote in oglegroups.com...
....
Lets try this. You are standing in the open with no wind. An
airplane passes by from left to right. If the airplane dropped a
cannon ball observers on the plane would see the ball drop straight
down, staying directly under the plane as it fell. The observer on
the ground would see the ball dropping from left to right. Following
its path back up leads to a point behind the current location of the
plane, where the plane was when it dropped the ball. This is
aberration. Just consider the airplane to be a stationary star and we
are on the earth moving in our orbit. Have the airplane fly back in
the opposit direction (we have continued our orbit under the
stationary star) and the aberration angle changes direction.

When the cannon ball was dropped a charge of black powder was lit off
with a bang, leaving a cloud of smoke in the stationary air at the
point where the ball was dropped. That is the point we will hear the
sound come from in our stationary air (dragged aether). The airplane
will have moved on in the time it took for the sound to reach us, so
the sound will come from behind the airplane, just like the cannon
ball. When the airplane flys back the other way the sound trails in
the opposite direction.

So far we agree.

If you want the star to be at rest in the aether of space just change
the airplane to a balloon floating back and forth in the jet streams.
It wont effect the final leg of the sound's passage to us in our
stationary air.

That's where we disagree. Rather than the jet stream, suppose
there is a uniform wind at all heights above 100m but the air
in that last 100m is still. There is a shear at that height.
Suppose someone on the balloon fires a gun to create a sound
wavefront.

The sound propagates vertically down through the air from
the balloon as seen by someone in the balloon and the
wavefront is always horizontal:

b
_|_
_|_
_|_
_|_
_|_
.... _|_ ....

From the point of view of someone on the ground, the ballon
and the sound waves are carried sideways by the wind but
the wavefronts remain horizontal. When the sound reaches
the shear, the balloon has drifte to 'b' from the point
where the gun was fired at 'g'

b g
_/_
_/_
_/_
_/_
_/_
.... _/_ ....

If there were a stationary observer at the bottom of the diagonal line
watching b drift by, where would he hear the sound come from, b or g ?

It appears to come from b because the direction is the
normal to the wavefronts. The same is true in the
completed diagram below so the aberration is the angle
xob rather than xog.

After the shear, the sound continues vertically and the
ballon drifts on to 'x' which it reaches when the gunshot
is heard on the ground by observer 'o':

x b g
_/_
_/_
_/_
_/_
_/_
.... _/_ .... shear
_|_
_|_
______o______ ground

The aberration is quite different from an aircraft
in uniformly moving air.

Yes, we use the down wind origin of the wave to calculate where the
wave front will be, but that is not where the sound came from.

It is where it appears to come from. Try drawing circles
radiating out from the source.

I disagree. This is what the real world experiment with sound
demonstrates. When there is a cross wind between two stationary
observers they still hear the sound come from the direction of the
source, not the down wind center of the wave front.

I have yet to be convinced of that. What was the link
to your experimental evidence again?

I provided no link. I was speaking of first hand observations you can
make yourself. Surely you have been in open areas when the wind was
blowing. Have you ever heard come from down wind of a stationary
object?

I have never tried it with equipment capable of measuring
the angle accurately enough and I doubt you have either. If
you have your eyes open, your brain identifies the source
with a combination of what your hear and what you see. Try
it wearing a blindfold and pointing to the source using
sound alone, get a friend to put two stakes in the ground
marking the direction, then take the blindfold off. You
will need a high wind speed to get a measurable offset.

I never have even when the distance was the better part of a
mile and the wind strong.

The distance doesn't matter, the sin of the angle is v/c
where c is the speed of sound.

When you view a car going by 40 feet away the angle defined by the car
is much larger than it is when the car is 4000 feet away. If the car
blows its horn at 40 feet and the sound is shifted back a foot it
would still seem to come from the front of the car. The same angle at
4000 feet would shift the sound by 100 feet, or about 5 car lengths
behind the car.

Yes, whether it is one foot in 40 or 100 in 4000, the
angle subtended at the listener is the same and it is
that angle that you measure so the distance is of no
interest in deciding whther you can measure the effect
by ear or if you need instrumentation. I don't believe
you could tell by ear without a very high wind speed.

I have often heard the sound from an
airplane come from a point behind its current location, so I can
detect a difference in direction if one exists.

An aircraft travels at 600 mph, I doubt you have tried
the experiment blindfold is as much as 60 mph when you
would get 1/10th of the displacement.

I have easily detected the shift with an old prop driven plane flying
at ~100 mph. I can also detect it when cars drive by at ~60 mph if I
am 100 yards or more from the highway. When a jet is traveling at 600
mph the effect is so great that it sometimes takes a few seconds to
find the jet.

I have also
experienced traveling at high rates of speed near others at race
tracks.

So are you saying that you have experienced travelling
at 60 mph parallel to another driver also moving at the
same speed and can definitely say that when he shouted
to you from the other car his voice did not appear to
come from behind his car by 5 degrees? See the diagram
below. I want to see some independent proof of that
claim before I accept it.

To determine the direction the sound is coming from the receiver needs
at least two points.

Right. Try this sketch:

S

--- wind

|
A--+--B

A and B are microphones and S is a source, say a gun which
emits a single spherical wavefront, or you can use a tone
generator and measure phase difference.

We are trying to duplicate a wave traveling in the aether. The
signals in the wires travel faster than the sound waves so you are in
effect using faster than light communication to determine the timing
of the reception at A and B.

But in a dragged aether, both the aether at ground level
and the cables are at rest relative to the ground (which
is how dragged aether explains the MMX) therefore the
speed of the signals in the cables is equal. As long as
the cables are the same length, there is no time
difference introduced, the actual speed is irrelevant.

snip as you did not comment

.. How can blocking the
line of sight path effect the path from down wind?

Because the sound is actually coming from the object which
hasn't moved but the wavefronts lie at an angle to the line
of sight like a swimmer crossing a river, he has to face
slightly upstream in order to swim perpendicular to the bank.

When that swimmer reaches you, after leaving the bank at a point
straight across, do you claim that he swam up from some point down
stream because that is where the water is that he swam through?

No, all I claim is that he is at the same angle to the
bank as he would have been had he swum from some point
downstream if the water was at rest. All we can measure
is the apparent source of the starlight.

George

In article . com,
"George Dishman" wrote:

On 6 Mar, 05:47, " wrote:
On Mar 3, 7:41 am, "George Dishman" wrote:
wrote in message
ps.com...
...
SR says there is no aberration
so there is no difference in the pointing angle. The predicted
pointing angle is the same but they disagree on whether there is
aberration due to the different definitions of aberration.

OK, but that's a hypothetical case. The Earth is in orbit
and what I am considering is whether a dragged aether can
the explain the actual observed aberration.

Lets try this. You are standing in the open with no wind. An
airplane passes by from left to right. If the airplane dropped a
cannon ball observers on the plane would see the ball drop straight
down, staying directly under the plane as it fell. The observer on
the ground would see the ball dropping from left to right. Following
its path back up leads to a point behind the current location of the
plane, where the plane was when it dropped the ball. This is
aberration. Just consider the airplane to be a stationary star and we
are on the earth moving in our orbit. Have the airplane fly back in
the opposit direction (we have continued our orbit under the
stationary star) and the aberration angle changes direction.

When the cannon ball was dropped a charge of black powder was lit off
with a bang, leaving a cloud of smoke in the stationary air at the
point where the ball was dropped. That is the point we will hear the
sound come from in our stationary air (dragged aether). The airplane
will have moved on in the time it took for the sound to reach us, so
the sound will come from behind the airplane, just like the cannon
ball. When the airplane flys back the other way the sound trails in
the opposite direction.

So far we agree.

If you want the star to be at rest in the aether of space just change
the airplane to a balloon floating back and forth in the jet streams.
It wont effect the final leg of the sound's passage to us in our
stationary air.

That's where we disagree. Rather than the jet stream, suppose
there is a uniform wind at all heights above 100m but the air
in that last 100m is still. There is a shear at that height.
Suppose someone on the balloon fires a gun to create a sound
wavefront.

The sound propagates vertically down through the air from
the balloon as seen by someone in the balloon and the
wavefront is always horizontal:

b
_|_
_|_
_|_
_|_
_|_
.... _|_ ....

From the point of view of someone on the ground, the balloon
and the sound waves are carried sideways by the wind but
the wavefronts remain horizontal.

***{Pardon me, but I can't resist butting in here.

Let's put some numbers on this. Suppose that there are two observers.
Observer A is standing on the ground. Above him is a stationary air mass
half a mile thick, and half a mile above that in an air mass moving to
the right at 100 mph there is a balloon containing observer B. For
simplicity, assume that the speed of sound is 770 miles/hour in both air
masses. If observer B fires a starter pistol when he is directly above
A, the resulting circular sound wave will expand outward. Its center
will move to the right with the balloon as the upper air mass itself
moves to the right, and one narrow segment of that circular wave will be
destined to eventually enter the ear of observer A. The question is,
which one?

As a first approximation, we are interested in the wave segment that
impacts the shear line at a position directly above A. We don't know how
long it will take to get to that position, so let's call that time t1.
When the wave segment in question impacts the shear line directly above
observer A, it will have traveled for t1 hours at 770 mph, and the
length of the radius along which it traveled will be 770 t1 miles.
Observer B will have moved to the right from his original position over
observer A by a distance of 100 t1 miles, and he will still be .5 miles
above the shear line. Thus we have a right triangle with a hypotenuse of
770 t1 and legs of 100 t1 and .5 miles. By the Pythagorean theorem, (770
t1)^2 = (100 t1)^2 + (.5)^2, and, solving, we find that t1 = .000655
hours, or 2.36 sec. Thus observer B has been displaced horizontally to
the right of A by a distance of 100(.000655) = .0655 miles, or 346 feet,
and the radial path taken by the wave segment from observer B to the
point on the shear line directly above observer A is 770(.000655) = .504
miles, or 2663 feet.

As the sound "ray" crosses the shear line, it will experience
aberration, and in this case we will have to use Sin^-1 (V1/V2), where
V1 is the component of motion of the receiver, the lower air mass,
perpendicular to the incoming "ray" of sound, and V2 is the speed of
sound. Therefore V1 = 100 cos [Cos^-1 (.5/.504)] = 100(.5/.504) =
100(.992) = 99.2 mph. Hence the aberration angle is Sin^-1 (99.2/770) =
7.4 degrees in the counterclockwise direction--i.e., toward the
vertical. And since the incoming ray was at an angle of Sin^-1 (100/770)
= 7.46 degrees clockwise away from vertical, it follows that the
aberration will approximately cancel out the original deflection of the
ray, causing it to point straight down in the lower air mass toward
observer A. Thereafter, it will travel straight down for a distance of
..5 miles at a speed of 770 mph, taking an additional time of t2 = .5/770
= .000649 hours, or 2.34 sec. Total transit time for the sound "ray"
will be t1 + t2 = 2.36 sec + 2.34 sec = 4.7 sec.

Let me emphasize that the above is an approximation. The problem here is
that observer B is only .5 miles above the shear line between the two
air masses, whereas in the case of stellar aberration the nearest star
is 93 million miles from the shear line between the Sun's pool and the
Earth's pool. Result: in the case of stellar aberration, we can treat
the incoming rays as if they are parallel, meaning that all the photons
impact the shear line between the Sun's pool and the Earth's pool at a
90 degree angle. Result: all have the same aberration angle to an
extremely high degree of accuracy as they cross the shear line. In the
present example, however, only the circular wave segment that is
directed straight down is going to impact the shear line at a 90 degree
angle. As we move away from that impact point, the angle varies
continuously, becoming smaller and smaller. Result: the component of
wind shear velocity perpendicular to the path taken by the incoming wave
segment (the sound "ray") becomes less and less, and the mathematically
simple solution is an approximation with less accuracy than is the case
when the same calculations are applied to stellar aberration.

In any event, I trust that my original point is now firmly established:
there is, in fact, no difficulty whatever in using the gravitationally
entrained aether theory to calculate aberration. The predictions done on
that basis do, in fact, match observation to a high degree of accuracy.
Moreover, the lower accuracy that results when applying the same
calculations to sound, as in the above example, are due to the fact that
the method used is, ultimately, an approximation that requires the
incoming rays to be parallel, rather than to any difficulty with the
underlying concepts. Wave aberration does, in fact, occur across the
shear line between moving media exactly as described above. The problem
associated with cases where the incoming rays cannot be treated as
parallel is mathematical, and applies to all theories of aberration,
whether they envision empty space, dragged aethers of various sorts, or
fully entrained aether.

--Mitchell Jones}***

[snip]

George

************************************************** ***************
If I seem to be ignoring you, consider the possibility
that you are in my killfile. --MJ

"Mitchell Jones" wrote in message
...
In article . com,
"George Dishman" wrote:

On 6 Mar, 05:47, " wrote:
On Mar 3, 7:41 am, "George Dishman" wrote:
wrote in message
ps.com...
...
SR says there is no aberration
so there is no difference in the pointing angle. The predicted
pointing angle is the same but they disagree on whether there is
aberration due to the different definitions of aberration.

OK, but that's a hypothetical case. The Earth is in orbit
and what I am considering is whether a dragged aether can
the explain the actual observed aberration.

Lets try this. You are standing in the open with no wind. An
airplane passes by from left to right. If the airplane dropped a
cannon ball observers on the plane would see the ball drop straight
down, staying directly under the plane as it fell. The observer on
the ground would see the ball dropping from left to right. Following
its path back up leads to a point behind the current location of the
plane, where the plane was when it dropped the ball. This is
aberration. Just consider the airplane to be a stationary star and we
are on the earth moving in our orbit. Have the airplane fly back in
the opposit direction (we have continued our orbit under the
stationary star) and the aberration angle changes direction.

When the cannon ball was dropped a charge of black powder was lit off
with a bang, leaving a cloud of smoke in the stationary air at the
point where the ball was dropped. That is the point we will hear the
sound come from in our stationary air (dragged aether). The airplane
will have moved on in the time it took for the sound to reach us, so
the sound will come from behind the airplane, just like the cannon
ball. When the airplane flys back the other way the sound trails in
the opposite direction.

So far we agree.

If you want the star to be at rest in the aether of space just change
the airplane to a balloon floating back and forth in the jet streams.
It wont effect the final leg of the sound's passage to us in our
stationary air.

That's where we disagree. Rather than the jet stream, suppose
there is a uniform wind at all heights above 100m but the air
in that last 100m is still. There is a shear at that height.
Suppose someone on the balloon fires a gun to create a sound
wavefront.

The sound propagates vertically down through the air from
the balloon as seen by someone in the balloon and the
wavefront is always horizontal:

b
_|_
_|_
_|_
_|_
_|_
.... _|_ ....

From the point of view of someone on the ground, the balloon
and the sound waves are carried sideways by the wind but
the wavefronts remain horizontal.

***{Pardon me, but I can't resist butting in here.

Let's put some numbers on this. Suppose that there are two observers.
Observer A is standing on the ground. Above him is a stationary air mass
half a mile thick, and half a mile above that in an air mass moving to
the right at 100 mph there is a balloon containing observer B. For
simplicity, assume that the speed of sound is 770 miles/hour in both air
masses. If observer B fires a starter pistol when he is directly above
A, the resulting circular sound wave will expand outward. Its center
will move to the right with the balloon as the upper air mass itself
moves to the right, and one narrow segment of that circular wave will be
destined to eventually enter the ear of observer A. The question is,
which one?

As a first approximation, we are interested in the wave segment that
impacts the shear line at a position directly above A. We don't know how
long it will take to get to that position, so let's call that time t1.

Let's take some slightly different numbers to make life easier.
Here's the full diagram:

x b g
_/_
_/_
_/_
_/_
_/_
.... _/_ .... shear
_|_
_|_
______o______ ground


Suppose the shear is 0.77 mile above the ground and the balloon
is 770 miles above that (thick atmosphere on this planet!). The
gun is fired when the balloon is 10 miles to the right of the
observer. The sound takes 1 hour to reach the shear during which
time the balloon drifts to the point directly above the observer.
The wavefront hits the shear parallel to its surface. It then
takes another 0.001 hour for the sound to reach the observer and
appears to have come form directly above. The balloon by then is
0.1 mile to the left.

Now consider the real numbers. A star is at least 4 light years
away or 120 million light seconds while the Earth-dragged aether
region must be much less than 1AU in diameter or at most 500 light
seconds.

When the wave segment in question impacts the shear line directly above
observer A, it will have traveled for t1 hours at 770 mph, and the
length of the radius along which it traveled will be 770 t1 miles.
Observer B will have moved to the right from his original position over
observer A by a distance of 100 t1 miles, and he will still be .5 miles
above the shear line. Thus we have a right triangle with a hypotenuse of
770 t1 and legs of 100 t1 and .5 miles. By the Pythagorean theorem, (770
t1)^2 = (100 t1)^2 + (.5)^2, and, solving, we find that t1 = .000655
hours, or 2.36 sec. Thus observer B has been displaced horizontally to
the right of A by a distance of 100(.000655) = .0655 miles, or 346 feet,
and the radial path taken by the wave segment from observer B to the
point on the shear line directly above observer A is 770(.000655) = .504
miles, or 2663 feet.

As the sound "ray" crosses the shear line, it will experience
aberration, and in this case we will have to use Sin^-1 (V1/V2),

No, that is what you are trying to calculate. The aberration is
the apparent displacement of the light from the source or angle
x-o-b. Clearly it will not be V1/V2 but is lessened by the ratio
of the distance travelled in the dragged aether to the distance in
the interstellar medium, or a factor of around a million depending
on your estimate of the reach of Earth's drag.

where
V1 is the component of motion of the receiver, the lower air mass,
perpendicular to the incoming "ray" of sound, and V2 is the speed of
sound.

To be precise the observer is at rest, it is the ballon that is
moving in the diagram above so it is the speed of the balloon
which is V1, but frames are equivalent so you could draw it like
this:

b b'
_|_
_|_
_|_
_|_
_|_
.... _|_ .... shear
_\_
_\_
______o______ ground


The apparent source is now b'.

George

On 9 Mar, 16:00, "George Dishman" wrote:

Suppose the shear is 0.77 mile above the ground and the balloon
is 770 miles above that (thick atmosphere on this planet!). The
gun is fired when the balloon is 10 miles to the right of the
observer. The sound takes 1 hour to reach the shear during which
time the balloon drifts to the point directly above the observer.

Of course that should have been ".. is 100 miles to the right of ..".

E&OE !!!

George

"Dan from Boston" wrote in message ...
Have any of these guys who are continually 'refuting' SR and GR ever taken
any math courses past algebra and trig? their analyses are pitiable.

Dan

Err.. yes. I wrote this one.
http://www.androcles01.pwp.blueyonde...tor/Vector.htm
Have you, pitiful Dan from Boston?

In article ,
(Dan from Boston) wrote:

Have any of these guys who are continually 'refuting' SR and GR ever taken
any math courses past algebra and trig? their analyses are pitiable.

***{Mainly, we have been discussing whether the gravitationally
entrained aether theory can or cannot explain stellar aberration, and
everyone who has posted anything on the topic, whether pro or con, has
made use of no math beyond trigonometry, for the simple and sufficient
reason that trigonometry is the math which such an analysis requires.

The only attack on relativity that was posted in connection with that
discussion was posted by me, and I assume from your comment that you
disagree with it, since you characterized my analysis as "pitiable." So
let me ask you a question: if someone told you that (a) automobile
speeds are a universal constant the value of which is 50 mph, and (b)
that the speed of each automobile has to be measured using an onboard
clock that automatically registers 1 hour for every 50 miles traveled,
would you accept his conclusion?

If not, then why would you accept Einstein's statement that (a) the
speed of light is a universal constant the value of which is 186,000
miles/sec, and (b) that the speed of light has to be measured using a
clock in the vicinity of the lightpath which automatically registers 1
hour for every 186,000 miles that light travels?

In other words, why can't we follow standard practice, and use clocks
calibrated to run at the same rate as standard time here on Earth?
That's what we do when we measure the speeds of automobiles and
everything else. Why must we make an exception for light?

Enquiring minds want to know! :-)

--Mitchell Jones}***

Dan

************************************************** ***************
If I seem to be ignoring you, consider the possibility
that you are in my killfile. --MJ